Question

a.  Unhealthy exhaust results from the combustion of octane (C8H10) in automobile engines via the following reaction.

C8H18(l) + 25O2(g)  16CO2(g) + 18H2O(g)

Assuming the gasoline is pure octane and that a suburban dweller burns about 6.44 x 104grams of octane (about 20 gallons) per week. How many grams of CO2 are produced (per week) to warm the globe? At 25oC and 1.00 atm, how many ml of CO2 are produced? Give your answer in scientific notation,

b.   During combustion, N2 and O2 in the air react to make the nasty brown gas NO2which is smog. The smog rises up into the atmosphere, combines with water vapor and rains down as nitric acid (HNO3) to acidify all the lakes and streams and kill the fish and plants. The following reactions are relevant.
A. O2(g) + N2 (g)  2NO(g)

B. 2NO(g) + O2(g)  2NO2(g)

C. NO2(g) + H2O(g)  HNO3(aq) + NO(g)

Balance reaction C. Calculate the moles of HNO3 that form if 2.2 x 10-4 moles of NO2rise up into the air and react completely with an unlimited amount of water vapor.

  c.    A 200 mL sample of the lake water is collected titrated directly with 0.01M NaOH and it takes only 3 mL of the NaOH titrant to reach the equivalence point. Calculate the molarity (M) of HNO3 in the lake water.

d. Nitric acid is a strong acid and engages in the following ‘proton’ transfer(dissociation) reaction with water (in which it is dissolved).

HNO3(aq) + H2O(l)  H3O+(aq) + NO3-(aq)

Suppose the concentration of HNO3 in the lakewater is 0.08 mg/mL. Calculate the molarity of H3O+ and the pH of the lakewater.

Answer 1

c) NaOH:

     M1 = 0.01 M

     V1 = 3 mL

   HNO3:

    M2 = ?

   V2 = 200 mL

WE know that

M1V1 = M2V2

M2 = M1V1/V2

     = 0.01 M x 3 mL/ 200 mL

    = 0.00015 M

M2 = 0.00015

Therefore,

molarity of HNO3 in lake water = 0.00015

d) [HNO3] = [H3O+] = 0.08 mg/ mL = 0.08 g/L

   [HNO3] in molarity = 0.08 g/L / molar mass of HNO3

                               = 0.08 g/L / 63 g/mol

                             = 0.00127 M

Hence,

[HNO3] = [H3O+] = 0.00127 M

Then,

pH = - log [H3O+]

     = - log [ 0.00127]

    = 2.9

Therefore,

pH of lake water = 2.9